Problem: The equation of hyperbola $H$ is $\dfrac {(y+7)^{2}}{4}-\dfrac {(x+3)^{2}}{16} = 1$. What are the asymptotes?
Solution: We want to rewrite the equation in terms of $y$ , so start off by moving the $y$ terms to one side: $\dfrac {(y+7)^{2}}{4} = 1 + \dfrac {(x+3)^{2}}{16}$ Multiply both sides of the equation by $4$ $(y+7)^{2} = { 4 + \dfrac{ (x+3)^{2} \cdot 4 }{16}}$ Take the square root of both sides. $\sqrt{(y+7)^{2}} = \pm \sqrt { 4 + \dfrac{ (x+3)^{2} \cdot 4 }{16}}$ $ y + 7 = \pm \sqrt { 4 + \dfrac{ (x+3)^{2} \cdot 4 }{16}}$ As $x$ approaches positive or negative infinity, the constant term in the square root matters less and less, so we can just ignore it. $y + 7 \approx \pm \sqrt {\dfrac{ (x+3)^{2} \cdot 4 }{16}}$ $y + 7 \approx \pm \left(\dfrac{2 \cdot (x + 3)}{4}\right)$ Subtract $7$ from both sides and rewrite as an equality in terms of $y$ to get the equation of the asymptotes: $y = \pm \dfrac{1}{2}(x + 3) -7$